How do you prove $1+tan^2 (x) = sec^2 (x)$ ?

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Answer 1 · 2016-05-02T21:37:32

Using the following:

for $cos(x)!=0$ , we have

$1+tan^2(x) = cos^2(x)/cos^2(x) + (sin(x)/cos(x))^2$

$=cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)$

$=(cos^2(x)+sin^2(x))/cos^2(x)$

$=1/cos^2(x)$

$=(1/cos(x))^2$

$=sec^2(x)$

How do you prove 1+tan^2 (x) = sec^2 (x)1+tan^2 (x) = sec^2 (x)?